Contents
Chapter-1
Drawing Instruments
Drawing instruments or drafting tools are used to draw neat and scaled drawings. The common drawing tools are listed as under:
1. Drawing board 2. Drawing sheet (A2) 3. Mini drafter
4. Set-squares (45o and 30o – 60o) 5. Drawing pencils (H and 2H) 6. Scale 30 cm
7. Master circle 8. French curve
9. Compass small and large size 10. Divider small and large size 11. Sketch book
12. Erase, knife and cello tape
Drawing Sheet (IS 10711:2001)
Standard size of the drawing sheets recommended by IS 10711:2001 are as follows:
Designation Dimension (mm)
A0 841*1189
A1 594*841
A2 420*594
A3 297*420
A4 210*297
A2 size sheet is preferred for class use
Sheet Layout (A2 size)
Spacing between border and trimmed edge 1. Top = 10mm
2. Right = 10mm 3. Bottom = 10mm 4. Left = 20mm
5. Title block size = 150x50mm Title Block (150mm X 50mm)
Rows height = 10 mm Column width = 50 mm Name:
NAME OF THE INSTITUTION Roll No:
Class
Title of Drawing Symbol
Scale
Title block Drawing Space
Trimmed Edge of A2 size Sheet Border
Sheet No. Date of Start Remark
Drawing Pencils
Depending upon the thickness of the line to be drawn pencils of different grades are used. The uses of drawing pencils of different grades are as follows:
H (medium grade) Used for drawing visible outlines and edges.
2H (Hard grade) Used for construction lines, dimension lines, leader lines, extension lines, center lines, etc.
Chapter 1
Technical Lettering:
Lettering is an important part of the Engineering drawing which provide the complete information about size of an object and appearance required. Writing of title, sub-title, dimensions and other relative details on drawing should be lettered with free hand. A good practice of free hand lettering improves the quality of drawing.
Normally two types of lettering are used by engineers, which are:
1. Single stroke letters 2. Block printing
Indian Standard Institution (I.S.I.) has recently been renamed as Bureau of Indian standard (B.I.S.) and adopted the International Standards Organization (l.S.O.) since 1983 as IS: 696-1972 to IS: 10714-1983. These standards are applicable to the following topics in the field of Engineering Drawings: -
(i) Drawing sheets (sizes, layouts etc.)
(ii) Lines: general principles of presentations of Technical Drawings.
(iii) Lettering of Technical Drawings (iv) Methods of dimensioning.
Single stroke lettering: Specimen letters and numerals as per IS:SP:46-1988
In single stroke lettering the thickness of the letter is uniform, it does not mean that the letter is made without lifting of the pencil. It seems to look that letters are drawn by single stroke of the pencil.
ABCDEFGHIJKLMNOPQRSTUVWXYZ
6-12ABCDEFGHIJKLMNOPQRSTUVWXYZ
10abcdefghijklmnopqrstuvwxyz abcdefghijklmnopqrstuvwxyz 1234567890 & 1234567890
10I II III IV V VI VII VIII IX X
10I II III IV V VI VII VIII IX X
106 8 6
6 8 6
Image
TYPES OF LINES
Engineering drawing uses various types of lines to describe different object and different purposes. Each type of lines in a drawing has definite meaning.
IS: 10714- 1983 adopted from ISO: 128-1982 specifies the types of lines and their applications as shown bellows:
Table
Dimension Technique:
It is the art of describing the size of an object by supplying complete information stating length, breadth, height, angle, size and position of holes etc. Element of dimensions are extension line, dimension line, leader line, arrowheads as shown in given figure.
Extension line: It is a continuous narrow line perpendicular to the outline to be dimensioned with
a gap of about 1mm from the outline.
Dimension line: It is a continuous narrow line drawn parallel to the edge of the part to be dimensioned. It can be placed inside or outside of the view. It is placed about 10mm away from the outline of the drawing.
Leader line: It is a continuous thin line connecting a dimensional value or a note with the corresponding feature on the drawing.
Dimension text: It is a numeral value shown along with the dimension line that is corresponding to the outline of the surface to be dimensioned.
Methods of Dimensioning:
Dimensioning of Circle and Arc
Circles of different size should be dimensioned and diameter is denoted by symbol as shown in the figure. Dimensions of small radii are usually shown outside the view while larger radii are shown inside the view.
Arcs should be dimensioned by their radii and denoted by 'R'. Only an arrow with its head on the arc end of the dimension line is used.
Dimensioning of Angles
Angular dimensions are oriented in different ways as shown in the following figure. Dimension text of small angles are placed outside the view whereas for large angles dimension text is placed inside the view.
Dimensioning Holes
Dimensioning of hole from the center line is not correct, except in the condition when the center line passes through the center of the hole. Center line has be extended so as to serve as an extension line.
The preferred way of to locate dimension the distance between the center of the holes is in the view in which the holes are visible. Also
System of placing dimension:
1. Aligned System 2. Unidirectional System
Aligned System of Dimensioning
Unidirectional System of Dimensioning
Arrangement of Dimensioning 1. Chain Dimensioning
2. Parallel Dimensioning
Scales:
In most of the cases drawing of big objects cannot be done in full size scale because the object is too big. On the other hand, drawing of the small objects also cannot be prepared in full size. Hence it is required to draw it with suitable scale as per the drawing sheet. So, the scale may be defined as "Ratio of the linear dimension of an element as represented in the drawing to the actual linear dimension of the same element". It is also known as draftsman scale.
Sizes of scale:
1. Full scale [ 1:1 ]
2. Enlarging scale [ 50:1, 20:1, 10:1, 5:1, 2:1 J 3. Reducing scale [ 1:2, 1:5, 1:10, 1:20, 1:50 ] Representative Fraction (R.F.):
The ratio of length on the drawing to the actual length in the same units. It is denoted by R.F.
R.F. = Length Actualof an lenthobject ofon thetheobjectdrawing `
For example: One cm long line on a drawing represents 50 meter length. So, R.F. = 1 cm/50 meters - 1/50 x 100 cm = 1/5000
R.F. = 1/5000 or 1:5000 Types of Scale:
1. Plain scale 1. Diagonal scale 2. Chord scale 3. Isometric scale 4. Vernier scale 5. Comparative scale Plane Scale:
It is used to represent either two units or one unit and its friction. A plain scale is a line, which is divided into suitable number of equal parts, the first part of which is sub- divided into smaller parts. :
Construction of Plain scale:
1. Calculate R.F., if not given 2. Calculate the length of scale
Length of scale = R.F. X maximum length of the scale 3. Divide the length of scale in equal number of part
4. Zero is placed at the end of first main division
5. This first division is farther sub-divided into equal number of parts
6. Numbering is done from zero mark, the units are numbered to the right and sub- divisions to the left
7. R.F. of the scale must be mentioned below the scale 8. Width of the scale is taken as 5mm.
Problem: Construct a plain scale such as to show meter when 1cm represents 6m and long enough to measure up to 60m. Find the R.F. and show a distance of 43 m.
Solution:
Given, Length of object on drawing - 1 cm Actual length of object = 6m = 600cm
R.F. = Length of object on drawing/Actual length of object
= lcm/6m= 1/600
= 1:600
Length of scale = R.F. X maximum length of scale L = (l/600) cm X 60m
= (1/600) X 6000 cm =10 cm.
Problem 1. Construct a plain scale of R.F. = 1:50000 to show kilometers and hectometers aid long enough to measure up to 9 kilometers. Measure a distance of 5 kilometers and 6 hectometers on the scale.
Problem 2. Draw a scale of 1:60 to show metes and decimeters and long enough measure upto 6 meters. Show a distance of 3.7 meters.
Problem 3. Construct a plain scale to show kilometers and hectometers when centimeters are equal to 1 kilometer and long enough to measure upto 6 kilometers. Find R.F. and show a distance of 4 kilometers and 5 hectometers on the scale.
Problem 4. The distance between Delhi and Aligarh is 132 kilometers. A train covers the distance in 2 hours and 12 minutes. Construct a plain scale to measure time upto a sing minute and mark a distance covered in 35 minutes. Take R.F. of the scale 1:400000.
Diagonal Scale:
It is used when measurements are required in three units or one unit and it friction upto second place of decimal points. In diagonal scale a line is divided into suitable number of equal parts, the first of which is sub-divided into smallest parts by diagonal. E.g. mm, cm, dm.
Problem: Construct a diagonal scale of a R.F. 1:500 to show meters and decimeters and long enough to measure upto 70 meters. Measure a distance 53.4 meters on the scale.
Solution: R.F. = 1:500 (given)
Length of scale = L = R.F. X Maximum length to be measured L = (1/500) x 70x 100 cm = 14 cm
Draw a line AB of 14 cm long
Maximum length to be measured is 70m, therefore, divide AB in 7 parts First main part is further divided into 10 equal parts
Draw a line AC 5cm long and perpendicular to line AB
Divide line AC in 10 equal parts and mark as 0,1,2,...,9 from A to C Join C to the first sub-division from A. thus the first diagonal is drawn Similarly, draw the remaining 9 diagonal
Complete the scale and show 53.4 meters on the diagonal scale.
Problem 1. Construct a diagonal scale of 3:200 showing meters, decimeters and centimeters and to measure upto 6 meters. Show a distance of 4.56 meter.
Problem 2. Construct a diagonal scale of R.F. 1/4000 to show meters and long enough to measure upto 500 meters. Show a distance of 354 meters.
Problem 3. Construct a diagonal scale of R.F. 1/50000 to show kilometers, hectometers and decameters and long enough to measure upto 6 kilometers. Measure a distance of 4 kilometers, 6 hectometer and 4 decameters.
Problem 4. The area of a field is 50000 sq. meters. The length and the breadth on the map are 10 cm and 8cm respectively. Construct a diagonal scale which can read upto 1 meter and long enough to measure upto 400m. Mark the length of 235 meter on the scale, also calculates the R.F. of the scale.
Chord Scale:
It is used for measuring and constructing angles when protector is not available. It is based on the length of the chords of different angles measured on the same arc.
Construction:
1. Draw a line AB of any length 2. Draw a perpendicular at B
3. Draw an arc of radius AB with centre B that cuts perpendicular at C 4. The arc AC represent an angle of 90° at the centre B
5. Divide AC in 9 equal parts, so each part represent an angle of 10°
6. Transfer each division point from the arc to the line AB by taking A as centre 7. Complete the scale by drawing a rectangle below AD
8. The division obtained are unequal
Problem: Construct an angle of 47° by means of chord scale.
Solution: Draw a line PQ
With any point O on it as centre and radius equal to 0-60 (from the chord scale), draw an arc cutting PO at a point A.
With A as centre and radius equal to 0-47 (chord of 47°) cut the arc at a point T.
Draw a line joining O with T, and then angle AOT is equal to 47°.
Problems on Plain & Diagonal Scale
1. A rectangular plot of 25 square kilometers is represented on a certain map by a similar rectangle of area 1 Sq. cm. Find the R.F.& draw a plain scale to show kilometers, indicate a distance of 55 km on it.
2. Construct a plain scale of 1:5 to show decimeter and centimeter and to read up to 1m. Show a length of 7.6 dm on it.
3. A rectangular plot of 6sq.km. is represented on a certain map by a similar rectangle of areal 1.0 sq.cm. Draw a plain scale to show units of 10 km. and single kilometer and long enough to read up to 60 km. Find the R.F. of the scale and show on it 63 meters.
4. Construct a plain scale of l cm=4km, to read kilometers and hectometers and long enough to read up to 6 kilometer. Find its R.F. and measure a distance of 65 km, 3 hm on the scale.
5. Draw a diagonal scale of R.F.= 3/100, showing meters, decimeters and centimeters and to measure up to 5m. Show a length of 5.69m on it.
6. An area of 144 sq.cm. on a map represents an area of 36sq km. on the field. Find the R.F. of the scale of the map and draw a diagonal scale to show kilometers, hectometers and decameters, and to measure up to 10km. Indicate a distance of 7km, 61im and 8dam
7. Distance of 300km. between Chandigarh and Delhi is represented by a line of 75mm. long on a railway map. What is the R.F. of the scale? Draw a scale showing single Kilometer and indicate on it a distance of 653km.
8. A map 120cm x 100cm represents an area of 3000 sq. km. Construct a diagonal scale to measure km., hm and dam. Find R.F. and indicate a distance of 7km 3hm 6dam.
9. Construct a diagonal scale to measure up to l/5th and l/25th of a centimeter. The length of the scale is 15cm.
10. Construct a diagonal scale of RF 1/500. It should be long enough to measure 100 meter. Show a distance of 64.4metre on the scale.
11. A rectangular plot of land of the area 45hectare is represented on a map by a similar rectangle of area 5sq. cm. Calculate the RF of the map. Also draw a diagonal scale to read up to single decameter, from the map. The scale should be long enough to read up to 5km. Show a distance of 3.47km. on the scale..
12. The distance between Bang lore and madras is 350 km. on a map it is represented by a line of length 70.4mm. What is the R.F. to which the map has been drawn?
Draw the diagonal scale of this R.F. to read up to one km., and long enough to measure 800 km. Mark on it a distance of 559 km and 217 km.
UNIT 2
Geometrical Construction Dividing Lines:
Let AB be the given line
Draw a line AC making an acute angle with AB (approx. 30o)
Divide the line AC in required numbers of equal divisions
Mark the points from A as 1,2,3,...N
Draw a line joining N to B :
Draw lines parallel to NB starting from 1,2,3.... To line AB
Then AT, 1'2', .... are the required divisions.
t-
Bisecting Lines and Arcs:
Let AB be the given line
With centre A and radius grater then half of AB draw arcs on both side of AB
With centre B and same radius draw arcs intersecting the previous arcs at C and D
Draw a line joining C and D that cuts AB at E
Then AE = EB = ½ of AB
Let AB be the arc drawn with centre O
Do the same as above
Then the line CD bisects the arc AB and passes through the centre O.
Draw Perpendicular lines:
Let AB be the given line and P the point in it. At which the perpendicular is required.
Construction:
With P as centre and any convenient radius, draw an arc cutting AB at C
With the same radius two equal divisions CD and DE are made
Again with the same radius and centers D and E, draw arcs intersecting each other at Q
Draw a line joining P and Q
Then PQ is the required perpendicular.
Draw Parallel lines:
Let AB be the given line and R the given distance Construction:
Mark P & Q as centers and radius equal to R, draw arcs on the same side of the AB
Draw the line CD, just touching the two arcs
CD is the required line.
Draw Tangent on Circle
With center O, draw the given circle and mark a point P on it
Draw a line joining O and P
Extend OP to Q so that PQ = OP
With centers O and Q draw arcs intersecting each other at R
Draw a line through P and R
Then this line PR is the required tangent.
Draw Tangent on Arc
Let AB be the given arc and P the point on it
With center P and any radius, draw arcs cutting the arc AB at C and D
Draw EF, the bisector of the arc CD it will pass through P
Through P, draw a line RS perpendicular to EF
Then this RS is the required tangent Construction of Plane Figures:
Square:
Draw a line AB equal to the given length
At A draw a line perpendicular to AB
With center A and radius AB, draw an arc cutting AE at D
With center D and B, with same radius, draw arcs intersecting at C
Draw lines joining C with B and D
Then ABCD is the required square.
Regular Polygons:
Draw a line AB equal to the given side of the polygon On AB as diameter, draw a semi-circle
With A as center and AB as radius, draw an arc on the same side
Draw a perpendicular bisector of AB cutting the semi- circle at 4 and arc at 6
Mark point 5 at the middle of 4 and 6
Mark point 7, so that 6-7 equals to 5-6, similarly point 8, 9 etc.
Heptagon:
With center 7 and radius A7, draw a circle Starting from B, cut it in 7 equal divisions with radius equal to AB
Draw lines BC. CD etc. and complete the pentagon
Similarly by taking 6 as center and A6 as radius hexagon can be drawn.
Conic Sections:
A cone is a surface generated by rotating a straight line such that it always keep contact with a closed curve called base, and contains a fixed point (apex) which does not lie in the plane of curve. When the axis of the cone is perpendicular on the base is called right circular cone. When a right circular cone is cut by section planes inclined to the axis at different angles, five different types of curves are obtained. These curves are called conic sections. These curves are circle, ellipse, parabola, hyperbola and isosceles triangle.
Circle:
When the cone is cut in such a way that the cutting plane is passed perpendicular to the axis and cuts all the generators of the cone, the section obtained is said to be circle.
Ellipse:
When the cone is cut in such a way that the cutting plane ‘1’ is inclined to the axis of the cone and cuts all the generators between apex and the base of the cone, the section obtained is said to be an ellipse. It is used in the construction of arches, bridges, base of utensils, pipe flanges, stuffing box, etc.
Parabola:
When the cone is cut in such a way that the cutting plane ‘2’ is inclined to the axis of the cone and parallel to one of the generators, the section obtained is said to be parabola. It is used for designing reflectors for parallel beams such as head lights of automobiles, solar concentrators, etc.
Hyperbola:
When the cone is cut in such a way that the cutting plane ‘3’ makes a smaller angle with the axis than that of the angle made by the generator of the cone, the section obtained is said to be hyperbola. It is used for designing cooling towers, hydraulic channels, etc.
Construction Methods of Ellipse 1. Eccentricity method
Construct an ellipse when the distance between the focus and the directrix is 50mm and the eccentricity is given as 2/3. Also draw the normal and tangent at any point P on the curve.
a. Draw a vertical line AB to represent directrix. At any point C on it draw a line perpendicular to directrix to represent the axis CC’.
b. Mark focus F on the axis at a distance of 50mm from the directrix.
c. Mark vertex V on the axis with the eccentricity 2/3 i.e. VF=20mm and VC=30mm.
d. Draw a line from focus F at 45o and extend CE, which intersect at K.
e. Drop a perpendicular from K on the axis that cuts it at point V’ i.e. second vertex.
f. Draw a line VE at point V on the axis equal to VF i.e. VE=VF.
g. Draw a line starting from C passing through E as shown in figure.
h. Along the axis CC’, mark 1, 2, 3, .... at approximately at equal distance.
i. Draw the lines normal to the axis CC’ from 1, 2, 3, …… upto the inclined line coming from C on both sides of the axis CC’ and give the name 1’, 2’, 3’, ….
respectively.
K
j. For point P1 and P1’, take a distance 1 1’ in the compass, F as center, mark on the line 1 1’ on both sides of the axis CC’ respectively.
k. Similarly for point P2 and P2’, take a distance 2 2’ in the compass, F as center, mark on the line 2 2’ on both sides of the axis respectively. Similarly for 3, 4,…..
l. Join all the points using French curves will give a closed curve i.e. an ellipse.
Normal and tangent on the curve at point P.
a. Take a point P on the curve as shown in the figure.
b. Make a line by joining P with F, and draw a line from F to directrix in such a way that FT is normal on PF.
c. Draw a line from T passing through P, this line is act as a tangent over the curve at point P.
d. Draw a line MN at point P normal PT, this line MN is act as normal to the curve at point P.
2. Four center method
Construct an ellipse by using four center method when the major and minor axis are given as 90mm and 50mm respectively.
a. Draw a line AB=90mm and CD=50mm in such a way that they are perpendicular to each other and intersecting each other at the mid of the lines at O.
b. Draw a line by joining AC.
c. Extend the line CD on both sides.
d. Draw an arc AK where O as center and OA as radius. Draw another arc where CK as radius and C as center that intersect line AC at L.
e. Bisect AL, this bisector intersect major and minor axis at M and N respectively.
f. Take the marks Q and R on the major and minor axis in such a way that OQ=OM and OR=ON respectively.
g. Draw the four arcs by using four centers M, N, Q and R in a way that M as center MA as radius, N as center NC as radius, Q as center QB as radius and R as center RD as radius.
h. The curve obtained by the arcs is an ellipse with major axis 90mm and minor axis 50mm.
3. Intersecting arcs method
Construct an ellipse when the major axis is 120mm and the distance between the foci is 108mm. Also determine the length of the minor axis.
a. Draw major axis AB=120 and bisect it by a line CD at point O.
b. Mark the foci F1 and F2 = 54 on both sides of the major axis.
c. Mark any number of points 1, 2, 3, …. At approximately equal distance on major axis in between the foci.
d. A1 as radius F1 and F2 as centers take marks on both sides of the major axis.
e. B1 as radius F2 and F1 as centers take marks on both sides of the major axis, which cuts the previous marks at point P1 as shown in figure.
f. Similarly for point P2, P3….., A2, A3 and B2, B3 will be the radii and F1 and F2 will be the centers.
g. Then draw a smooth curve through points P1, P2, P3,……., this curve will be an ellipse by using intersecting arc method.
4. Concentric circle method
Draw an ellipse using concentric circle method given that the major axis is 120mm and minor axis is 80mm.
a. Draw a major axis AB=120mm and minor axis CD=80mm in such a way that they are mutually perpendicular and intersecting each other at the mid of the axis at O.
b. Draw two concentric circle i.e. major axis circle and minor axis circle, O as center and 60mm and 40mm as radii respectively.
c. Divide the circle into 12 equal parts. Give the name 1, 2, 3, 4,……., 12 and 1’, 2’, 3’, 4’, ………, 12’.
d. Draw the radial lines joining O to 1, 2, 3, 4, ……, 12.
e. From 1 draw a vertical line parallel to CD, and from 1’draw a horizontal line parallel to AB. Both intersect at P1.
f. Repeat the above and obtained all the 12 points i.e. P2, P3, P4,……, P12.
g. Draw a smooth curve through these point. This curve is an ellipse generated by the concentric circle method.
5. Rectangle method
Draw an ellipse using rectangle method given that the major axis is 120mm and minor axis is 80mm.
a. Draw a major axis AB=120mm and minor axis CD=80mm in such a way that they are mutually perpendicular and intersecting each other at the mid of the axis at O.
b. Draw a rectangle EFGH of 120mm x 80mm, in such a way that O is the center of the rectangle as shown in the figure.
c. Divide OA and AE in equal number of parts say 4.
d. Draw three line by joining C to 1’, 2’, and 3’.
e. Draw a line starting from D passing through 1 and intersect line C1’at P1.
f. Repeat the same for point P2 and P3.
g. Transfer these points on the other sides of the axis to get rest of the points.
h. Draw a smooth curve through these point. This curve is an ellipse generated by the rectangle method.
Construction Methods of Parabola 1. Eccentricity method
Construct a parabola when the distance between the focus and the directrix is 40mm and the eccentricity is given as 1. Also draw the normal and tangent at any point P on the curve.
a. Draw a vertical line DD to represent directrix. At any point C on it draw a line perpendicular to directrix to represent the axis CA.
b. Mark focus F on the axis at a distance of 40mm from the directrix.
c. Mark vertex V on the axis with the eccentricity 1 i.e. VF=20mm and VC=20mm.
d. Draw a line VS at point V on the axis equal to VF i.e. VS=VF.
e. Draw a line at 45o starting from C passing through S as shown in figure.
f. Along the axis CA, mark 1, 2, 3, .... at approximately at equal distance.
g. Draw the lines normal to the axis CA from 1, 2, 3, …… upto the inclined line coming from C on both sides of the axis CA and give the name 1’, 2’, 3’, ….
respectively.
h. For point P1 and P1’, take a distance 1 1’ in the compass, F as center, mark on the line 1 1’ on both sides of the axis CA respectively.
i. Similarly for point P2 and P2’, take a distance 2 2’ in the compass, F as center, mark on the line 2 2’ on both sides of the axis respectively. Similarly for 3, 4,…..
j. Join all the points using French curves will give a curve i.e. a parabola.
Normal and tangent on the curve at point P.
e. Take a point P on the curve as shown in the figure.
f. Make a line by joining P with F, and draw a line from F to directrix in such a way that FQ is normal on PF.
g. Draw a line from Q passing through P, this line is act as a tangent over the curve at point P.
h. Draw a line MN at point P normal PQ, this line MN is act as normal to the curve at point P.
2. Rectangle method
Construct a parabola by using rectangle method with a base of 100mm and axis 60mm.
a. Draw AB = 100mm to represent the double ordinate. Bisect AB at E.
b. Draw an axis EF of 60mm at point E in a way that EF is normal to AB.
c. Draw a rectangle ABCD as shown in figure.
d. Divide AE and AD in equal number of parts say 4.
e. Draw lines by joining F with 1, 2 and 3.
f. For point P1, draw a line starting from 1’ parallel to EF that intersect line F1.
This point of intersection gives point P1.
g. Repeat the same for points 2 and 3.
h. Transfer these points on the other side of the ordinate EF.
i. Join all the points starting from A passing through F and reaches to B.
3. Tangent method
A water jet discharges water from ground level at an inclination of 50o to ground.
The jet travels a horizontal distance of 9m from the point of discharge. Trace the path of the jet.
a. Select a scale 1:100, therefore draw the horizontal travel PQ=90mm.
b. As the jet is 50o from the ground, therefore draw two lines from P and Q inclined at 50o from the line PQ. These lines will intersect at R.
c. Bisect PR at O and draw a normal at O which passes through R. and also bisect OR at V.
d. Divide PR and QR in equal number of parts say 8, i.e. 1, 2, 3, …. and 1’, 2’, 3’,
….. respectively as shown in the figure.
e. Draw the tangents by joining 1 with 1’, 2 with 2’, similarly upto 7 with 7’.
f. Draw a smooth curve in such a way that the above lines are act as the tangent over the curve.
Construction Methods of Hyperbola 1. Eccentricity method
Construct a hyperbola when the distance between the focus and the directrix is 40mm and the eccentricity is given as 4/3. Also draw the normal and tangent at any point P on the curve.
a. Draw a vertical line DD to represent directrix. At any point C on it draw a line perpendicular to directrix to represent the axis CA.
b. Mark focus F on the axis at a distance of 40mm from the directrix.
c. As eccentricity e is given as 4/3, therefore construct a right angled triangle CXY, where XY is 4units and CX is 3units.
d. Draw a line starting from F and at 45o from CF, that intersect line CY at S.
e. Draw a vertical line from S, that cuts the axis at V i.e. vertex of the curve.
f. Along the axis CA, mark 1, 2, 3, .... at approximately at equal distance.
g. Draw the lines normal to the axis CA from 1, 2, 3, …… upto the inclined line coming from C on both sides of the axis CA and give the name 1’, 2’, 3’, ….
respectively.
h. For point P1 and P1’, take a distance 1 1’ in the compass, F as center, mark on the line 1 1’ on both sides of the axis CA.
i. Similarly for point P2 and P2’, take a distance 2 2’ in the compass, F as center, mark on the line 2 2’ on both sides of the axis respectively. Similarly for 3, 4,…..
j. Join all the points using French curves will give a curve i.e. a hyper parabola.
Normal and tangent on the curve at point P.
a. Take a point P on the curve as shown in the figure.
b. Make a line by joining P with F, and draw a line from F to directrix in such a way that FQ is normal on PF.
c. Draw a line from Q passing through P, this line is act as a tangent over the curve at point P.
d. Draw a line MN at point P normal PQ, this line MN is act as normal to the curve at point P.
2. Rectangle method
Construct a rectangular hyperbola when a point P on it is at a distance of 20mm and 30mm respectively from the two asymptotes.
a. Draw OA and OB in a way that OB is normal at OA, i.e. <BOA=90o. b. Mark F on OA where OF=20mm.
c. Mark C on OB where OC=30mm.
d. Draw CD parallel to OA and FE parallel to OB, which intersects at point P.
e. Take any number of points on CD say 1, 2, 3, …….., 6.
f. Draw the lines by joining O to 1, O to 2, ……. O to 6.
g. These lines intersect EF, give name to all the intersecting points corresponding to the points name given on the CD i.e. 1’, 2’, 3’, …….., 6’.
h. For point P1, draw a vertical line from 1 and a horizontal line from 1, the intersection of these lines will give the point P1.
i. Repeat the same for points P2, P3, ….., P6.
j. Join all the points with a smooth curve, this curve is a rectangular hyperbola.
Involutes:
The involutes is a curve traced out by an end of a piece of thread unwound from a circle or a polygon. Involutes of a circle is used as teeth profile of the gear wheel.
Types of involutes:
1. Involutes of a circle
An inelastic string of length 132mm is wound around a circle of 42mm diameter. Draw the path traced by the end of the string. And also draw normal and tangent at any point N on the curve.
a. Draw a circle of diameter 42mm with a center O.
b. Divide the circle into 12 equal parts of 30o each. And give the name 1, 2, ….., 12, as shown in figure.
c. Draw tangents at the points on the circle i.e. from 1, 2, 3, ……, 12.
d. Extend the tangent at the point 12 and mark PQ on it such that PQ=132mm.
e. Divide PQ in 12 equal parts and mark 1’, 2’, 3’, …… starting from P.
f. For point P1 take a distance in the compass P1’ and 1 as center cut tangent 1.
g. Repeat the same upto point P12 i.e. Q.
h. Join all the points with a smooth curve, this curve is an involute of a circle of diameter 42mm.
To draw a tangent and normal on the curve at point N.
a. Take a point N on the curve, and join N with O.
b. Bisect line ON and mark as X.
X
c. Draw an arc of radius XN and X as center that cuts the circle at point M.
d. Join M with N, this line MN will be the normal on the curve at point N.
e. Draw a line ST perpendicular to the line MN, this line ST is a required tangent at point N on the curve.
2. Involutes of a polygon
Draw the involute of an equilateral triangle of side 24mm.
a. Draw an equilateral triangle ABC of side 25mm.
b. Extend line CA to P1 in such a way that AP1=AB.
c. Extend line BC to P2 in such a way that CP2=CP1.
d. Extend line AB to P3 in such a way that BP3=BP2.
e. With A as center AB as radius draw a curve BP1.
f. With C as center CP1 as radius draw a curve P1P2.
g. With B as center BP2 as radius draw a curve P2P3.
h. This B,P1,P2,P3 represents the involute of the given triangle.
Draw the involute of a square of the side 25mm. draw a tangent and normal at any point M.
a. Draw square ABCD of side 25mm
b. Extend line DA to P1 in such a way that AP1=AB.
c. Extend line CD to P2 in such a way that DP2=DP1.
d. Extend line BC to P3 in such a way that CP3=CP2.
e. Extend line AB to P4 in such a way that BP4=BP3.
f. With A as center AB as radius draw a curve BP1.
g. With D as center DP1 as radius draw a curve P1P2.
h. With C as center CP2 as radius draw a curve P2P3.
i. With B as center BP3 as radius draw a curve P3P4.
j. This B, P1, P2, P3, P4 represents the involute of the given square.
To draw a tangent and normal on the curve at point M.
a. Take any point M on the curve, such that the point M lies on the arc between P3P4.
b. Draw a line by joining M with the center of the arc P3P4 i.e. point B.
c. This line MN is the required normal on the curve.
d. At M draw a perpendicular to the normal to obtained the tangent TT
Cyclonical Curves:
These curves are generated by a fixed point on the circumference of a circle, which rolls without slipping along a fixed straight line or a circle. The rolling circle is called
generating circle and the fixed straight line or circle is called directing line or directing circle.
Cycloid:
It is a curve generated by a fixed point on the circumference of a circle, which rolls without slipping along a fixed straight line.
A circle of 42mm diameter rolls over a flat surface without slipping. Draw the path trace by the point P present on the circumference of the circle for one complete revolution. Also draw the tangent and normal at any point on the curve.
a. Draw a circle of diameter 42mm with center C.
b. Divide the circle into 12 equal parts of 30o each, and give the name 1’, 2’, 3’, …..
12’, starting from P anticlockwise.
c. Draw a horizontal line from P, equal to the circumference of the generating circle.
That is PA=*D. PA=(22/7)*42=132mm
d. Divide line PA in 12 equal number of parts, and give the 1, 2, 3,……. 12, starting from P.
e. Draw line from 1’, 2’, 3’,… parallel to line PA.
f. Line CB parallel and equal to PA is divided in to 12 equal parts, and give the name C1, C2, C3,…… C12, starting from C.
g. For point P1, R as radius C1 as center take a mark on the horizontal line coming from Point 1’.
h. Similarly for point P2, R as radius C2 as center take a mark on the horizontal line coming from Point 2’. Repeat the same for all the points P3, P4, …… P12.
i. Obtain the curve by joining all the points. This curve is the cycloid of the given circle.
To draw normal and tangent on the curve at point N.
a. Take a point N on the curve.
b. Take a distance R in the compass N as center mark on the center line CB at M.
c. Draw a vertical line from M upto line PA that cuts at O.
d. Join O with N, the obtained line ON is the required normal at point N on the curve.
e. At N draw a perpendicular to the normal ON to obtain the required tangent ST.
Epicycloids:
It is a curve generated by a fixed point on the circumference of a circle, which rolls without slipping along the out-side circumference of another fixed circle.
Draw an epicycloid of rolling circle 40mm diameter (2r), which rolls over a base circle of diameter 150mm (2R) for one complete revolution. Also draw tangent and normal at any point on the curve.
.
a. As shown in the figure the generating circle of radius 20mm rolls one complete revolution over the directing or base circle of radius 75mm, therefore the arc PQ
is equal to the circumference of the generating circle. i.e. arc PQ= 2r. So that the arc angle = (r/R)*360o, i.e. = 96o.
b. Take a point O as center and R=75mm as radius, draw an arc PQ with angle 96o. c. Let P be the generating point. Extend line OP to C where PC= r = 20mm.
d. Draw the generating circle where C as center and CP as radius.
e. Divide the generating circle and the arc PQ in equal number of parts say 12.
f. Mark 1, 2, 3, … 12 on the generating circle and C1, C2, C3, ……., C12 on the on the arc coming from the center of the circle C.
g. Draw the arcs with center O and radius O1, O2, O3, ……, O12.
h. For point P1, C1 as center r=20mm as radius take a mark on the arc coming from the point 1.
i. For point P2, C2 as center r=20mm as radius take a mark on the arc coming from the point 2. Repeat the same for the points P3, P4, ….. P12.
j. Join all the points with a smooth curve, this curve is said to be an epicycloid.
To draw normal and tangent on the curve at point N.
a. Take a point M on the curve.
b. Take a distance r=20mm in the compass, M as center mark on the central arc at N.
c. Draw a radial line from N to the center of arc O that cuts the directing arc at S.
d. Join S with M, the obtained line SN, which is the required normal at point M on the curve.
e. At M draw a perpendicular to the normal SN to obtain the required tangent TT.
Hypocycloid:
It is a curve generated by a fixed point on the circumference of a circle, which rolls without slipping along the in-side circumference of another fixed circle.
Draw a hypocycloid of a circle of 40mm diameter which rolls inside of a circle of 200mm diameter for one complete revolution. Also draw normal and tangent on the curve at any point N.
a. As shown in the figure the generating circle of radius 20mm rolls one complete revolution inside a directing or base circle of radius 100mm, therefore the arc PA is equal to the circumference of the generating circle. i.e. arc PA= 2r. So that the arc angle = (r/R)*360o, i.e. = 72o.
b. Take a point O as center and R=100mm as radius, draw an arc PA with angle 72o. c. Let P be the generating point. Take a point C on line OP where PC= r = 20mm.
d. Draw the generating circle where C as center and CP as radius.
e. Divide the generating circle and the arc PA in equal number of parts say 12.
f. Mark 1, 2, 3, … 12 on the generating circle and C1, C2, C3, ……., C12 on the on the arc coming from the center of the generating circle C.
g. Draw the arcs with center O and radius O1, O2, O3, ……, O12.
h. For point P1, C1 as center r=20mm as radius take a mark on the arc coming from the point 1.
i. For point P2, C2 as center r=20mm as radius take a mark on the arc coming from the point 2. Repeat the same for the points P3, P4, ….. P12.
j. Join all the points with a smooth curve, this curve is said to be an epicycloid.
To draw normal and tangent on the curve at point N.
a. Take a point N on the curve.
b. Take a distance r=20mm in the compass, N as center mark on the central arc at D.
c. Draw a radial line from D to the directing arc that cuts the directing arc at M.
d. Join M with N, to obtain the line MN, which is the required normal at point N on the curve.
e. At N draw a perpendicular to the normal MN, to obtain the required tangent ST.
Helix:
It is defined as a curve generated by a point, moving around the surface of a right circular cylinder with uniform angular velocity and with uniform linear velocity in axial direction of the cylinder. The axial advance of the point during one complete revolution is called pitch of the helix.
Draw a helix given that the cylinder diameter is 50mm and pitch is 75mm.
a. Draw the front and top view of the given cylinder.
b. Divide the top view i.e. a circle into 12 equal number of parts in clockwise direction, and give the name 1, 2, 3, ….., 12. As shown in the figure.
c. Divide the pitch into same number of parts i.e. 12 and say 1’, 2’, 3’, ….., 12’.
d. Draw a vertical from point 1 present on the circle and a horizontal from the point 1’
present on the pitch length. The intersection will give the point P1.
e. Repeat the same for point P2, P3, …… P12.
f. Join all the points from P to P12 with a smooth curve. The obtained curve is the required helix.
Spirals:
If a line rotates in a plain about one of its ends and at the same time, a point moves along the line continuously in one direction the curve traced out by the moving point is called spiral. The point about which the line rotates is called pole.
If a curve is traced out by a point moving in such a way that its movement towards or away from the pole is uniform with the increase of the vectorial angle from the starting line is called Archimedean spiral.
A point P is 120 mm away from the fixed point pole O. A point P moves towards pole O and reaches the position Q in one convolution where OQ is 24 mm. The point P moves in such a way that its movement towards fixed point O, being uniform with its movement around fixed point pole O. Draw the curve traced out by the point P.
Procedure:
1. With pole O as centre and radii equal to R1 = 120 mm and R2 = 24 mm draw two circles.
2. Divide 360o at pole O into twelve equal parts and draw radial lines OP, O1’, O2’…, O12’.
3. Divide (R1 – R2) = 120 – 24 = 96 mm, length (PQ=96mm) also into same twelve equal parts as shown in the figure.
4. Initial position of point is P.
5. Point P 1 is obtained by cutting radial line O1’ by arc with centre O and having radius as O–11 length.
6. Similarly, points P2, P3, .. etc. are obtained.
7. Observe that point P 12 is coincident with Q.
8. Thus, point P starts moving towards pole O and reaches the point Q in one convolution and hence the curve obtained by joining P, P 1 , P2 ,… P 12 in proper sequence is an Archimedean Spiral.
Problems based on Geometrical Curves:
Ellipse
To construct an ellipse when the distance of the focus from the directrix is equal to 50 mm and eccentricity is 2/3.
Parabola
1. Draw a parabola, given the distance between focus F and directrix DD" as 50 mm.
Also draw a tangent to the curve at appoint on it.
2. . Draw a parabola by tangent method given its base AB as 90 mm and the distance CD from AB to the vertex D as 75 mm.
3. To construct a parabola, when the distance of the focus from the directrix is 50mm.
Hyperbola
1. Construct a hyperbola, when the distance of the focus from the directrix is 65mm and eccentricity is 3/2.
2. Draw a rectangular or equilateral hyperbola, given the position of point P on it as 25mm and 20mm from the asymptotes OX and OY, respectively. Also draw a tangent and a normal to it at a given point M on it.
Involutes
Draw an involute to a circle of diameter 50mm.
Draw an involute of a given square of side 4mm.
Draw involutes of the plane figures given below.
Cycloidal Curves
1. To construct a cycloid, given the diameter of the generating circle.
2. Draw an epicycloids, given the radii of rolling & directing circles as r=30mm and R= 120mm, respectively. Also draw a normal and a tangent at any point on the curve.
3. Draw a hypocycloid when radius of the directing circle R=60mm and radius of the rolling circle r=30 mm.
Spiral & Helix
1. Construct an Archimedean Spiral of one convolution given the radial movement of the point P during one convolution as 60 mm and the initial position of P as pole.
2. Construct a Logarithmic spiral of one convolution, given the length of the shortest radius as 21 mm and the ratio of the length of the radius vectors enclosing an angle of 30° as 8/7.
3. Draw a single start helix of 80 mm pitch, on a vertical cylinder of diameter 50 mm.
Also develop the helix.
4. Draw a helix of 72 mm pitch of one revolution around a right circular cone of diameter 60 mm & height 72 mm. Also show the plan of the helix.
UNIT III
ORTHOGRAPHIC PROJECTION
It is required to provide each and every information about the shape and size of any 3D object to produce in any manufacturing industry. This work can only be done by using orthographic projections. It is a means of representing 3D objects on the two dimensional plane. It is a form of parallel projections, in which all the projection lines are at right angle to the projection plane.
This concept was invented by a French scientist Monge in 1800. The projection is a Latin word means through forward. The term orthographic is sometimes reserved specifically for depictions of objects where the principal axis or planes of the object are also parallel with the projection plane, but these are better known as multiview projections. Furthermore, when the principal planes or axis of an object in an orthographic projection are not parallel with the projection plane, but are rather tilted to reveal multiple sides of the object, the projection is called an axonometric projection.
For projecting an image on the plane straight lines are drawn from various points of the object to meet the transparent plane at right angle. Thus the object is said to be projected on that plane of projection.
Plane of Projection: the transparent plane on which the projections are drawn in known as plane of projection.
View: The image formed on the plane of projection is called view.
Projection: The view formed by joining the points in the correct sequence at which rays meet the plane is called the projection of the object.
Projectors: The lines or rays drawn from the object to the plane are called projectors.
First Angle and Third Angle Projections [BIS CONVENTIONS (IS 15021:2001)]
Four Quadrants: When the planes of projection are extended beyond the intersection line, four quadrants are formed. These quadrants are named as I, II, III and IV quadrants in the counter clockwise direction as shown in figure. To open out the plane in 2D frame, only horizontal plane rotates in clockwise direction to open it, therefore for orthographic projections only First and Third quadrant can be used, second and fourth quadrant cannot be used.
The observer will always be on the right of the four quadrant and facing the vertical plane. The position of the object is defined in respect to the observer and the plane i.e. in front of the VP or behind the VP, above the HP or below the HP.
First Angle of Projection: In this the object lies in front of the vertical plane and above the horizontal plane i.e. in the First quadrant. Concept of first angle of projection is as follows:
Object lies between the observer and the VP and also above the HP.
Projectors of the object lies on the VP to form the front view and on the HP to form top view.
The quadrant always opens by rotating the HP in clockwise direction.
By opening the quadrant the top view comes below the front view i.e. XY or reference line and the front view comes above the XY line.
The symbol used for the first angle of projection is shown in the figure.
Third Angle of Projection: In it the object is placed behind the vertical plane in respect to the observer and below the horizontal plane i.e. in the third quadrant. The front view comes below the XY line and top view above the XY line. The symbol used for the third angle of projection is shown in the figure.
Projection of Point
A solid is formed by three dimensions, viz., length, width and height. If one of the dimension is made zero, the solid becomes a 2D plane. If two out of three dimensions made zero, then it becomes a line. If all the three dimensions converted to zero then the solid becomes a point. A point may lie in space, in any of the four quadrants formed by two reference planes horizontal plane and vertical plane. It is represented by a dot (.) in the drawings.
Point in First Angle of Projection
Prob. 1. Draw the projections of Point A is 30 mm above Horizontal Plane and 45 mm in front of Vertical Plane.
Sol. In the given problem the point A lies in the first angle of projection.
Mark a point a’ 30mm above the XY in the V.P. and a 45mm in-front of the V.P. in the H.P.
Projection of Point
1. A point A is 25 mm above the H.P. and 30 mm infront of the V.P. Draw its projection
2. A point A is 20mm below the H.P. and 30 mm behind the V.P. Draw its projection 3. Draw the projection of the following points on the same ground line, Keeping the
projectors 25 mm apart:
A, in the H.P. and 20 mm behind the V.P.
B, 40 mm above the H.P. and 25 mm in front of the V.P C, in the V.P. and 40 mm above the H.P.
D, 24 mm below the H.P and 25 mm behind the V.P.
E, 15 mm above the H.P. and 50 mm behind the V.P.
F, 40 mm below the H.P. and 25 mm in front of the V.P G, in both the H.P. and the V.P.
Projection of line
1. A line PQ, 90 mm long, is in the H.P. and makes an angle of 30° with the V.P. Its end P is 25 mm infront of the V.P. Draw its projections.
2. The length of the top view of a line parallel to the V.P. and inclined at 45° to the H.P. is 50 mm. One end of the line is 12mm above the H.P and 25 mm infront of the V.P Draw the projections of the line and determines its true length.
3. Draw the projection of a 75 mm long straight line, in the following positions:
(i) Parallel to both the H.P. and the V.P. and 25mm from each, (ii) Parallel to and 30 mm above the H.P. and in the V.P.
(iii) Parallel to and 40 mm in front of the V.P. and in the H.P.
(iv) Perpendicular to the H.P., 20 mm in front of the V.P. and its one end is 15 mm above the H.P.
(v) Perpendicular to the V.P., 25mm above the H.P. and its one end in the V.P.
(vi) Perpendicular to the H.P., in the V.P. and its one end in the H.P.
(vii) Inclined at 30° to the H.P. and its one end 20 mm above it; parallel to and 30 mm in front of the V.P.
(viii) Inclined at 60° to the V.P. and its one end 15 mm infront of it; parallel to and 25 mm above the H.P.
4. A point A is 50 mm below the H.P. and 12 mm behind the V.P. A point B is 10 mm above the H.P. and 25 mm in front of the V.P. The distance between the projectors of A and B is 40 mm. Determine the traces of the line joining A and B.
5. A line AB, 50 mm long, has its end A in both the H.P. and V.P. It is inclined at 30° to the H.P. and at 45° to the V.P. Draw its projections.
6. The top view of a 75 mm long line AB measures 65 mm, while the length of its front view is 50 mm. Its one end A is in the H.P. and 12 mm in front of the V.P.
Draw the projections of AB and determine its Inclinations with the H.P. and the V.P.
7. A line AB, 65mm long, has its end A 20mm above the H.P. and 25mm in front of the V.P. The end B is 40 mm above the H.P. and 65 mm in front of the V.P.
Draw the projections of AB and show its inclinations with the H.P. and the V.P.
8. A line AB, 90 mm long, is inclined at 45' to the H.P. and its top view makes an angle of 60° with the V.P. the end A is in the H.P. and 12 mm in front of the V.P.
Draw its front view and find its true inclination with the V.P.
Projection of planes
1. An equilateral triangle of 50 mm side has its V.T. parallel to and 25 mm above xy.
It has no H.T. Draw its projections when one of its sides is inclined at 45° to the V.P.
2. A square ABCD of 40 mm side has a corner on the H.P. and 20 mm in front of the V.P. All the sides of the square are equally inclined to the H.P. and parallel to the V.P. Draw its projection and show its traces.
3. A regular pentagon of 25 mm side has one side on the H.P. Its plane is inclined at 45° to the H.P. and perpendicular to the V.P. Draw its projections.
4. A square ABCD of 50 mm side has a corner A in the H.P., its diagonal AC inclined at 30° to the H.P. and the diagonal BD inclined at 45° to the V.P. and parallel to the H.P. Draw its projections.
5. Draw the projection of a circle of 50mm. diameter resting in the H.P. at a point A on the circumference; its plane inclined at 45° to the H.P.
(a) the top view of the diameter AB making 30° angle with the V.P.;
(b) the diameter AB making 30° angle with the V.P.
Projection of solids
1. Draw the projections of a triangular prism, base 40mm side and axis 50mm long, resting on one of its base on the H.P with a vertical face Perpendicular to V.P.
2. Draw the projections of a pentagonal pyramid, base 30mm edge and axis 50mm long, having its base on the H.P and edge of the base parallel to the V.P. Also draw its side view.
3. Draw the projections of a pentagonal prism, base 25mm side and axis 50mm long, resting on one of its rectangular faces on the H.P with the axis inclined at 45° to the V.P.
4. Draw the projections of a cylinder 75mm diameter and 100mm long, lying on the H.P with its axis inclined at 30° to the V.P. and parallel to the H.P.
5. Draw the projections of a cone, base 45mm diameter and axis 50mm long, when it is resting on the H.P. on a point on its base circle with (a) the axis making an angle of 30° with the H.P. and 45° with the V.P.; (b) the axis making an angle of 30°
with the H.P. and its top view making 45° with the V.P.
6. A pentagonal pyramid, base 25mm side and axis 50mm long has one of its
triangular faces in the V.P. and the edge of the base contained by that face makes an angle of 30° with the H.P. Draw its projections.
Sections of solid
1. A pentagonal pyramid, base 30mm side and axis 65mm long, has its base horizontal and an edge of the base parallel to the V.P. A horizontal section plane cuts it at a distance of 25mm above the base. Draw its front view and sectional top view.
2. A hexagonal pyramid, base 30mm side and axis 75mm long, resting on its base on the ground with two of its edges parallel to the V.P. is cut by two section planes, both perpendicular to the V.P. The horizontal section plane cuts the axis at a point 35mm from the apex. The other plane, which makes an angle of 45 with the H.P., also intersects the axis at the same point. Draw the front view, sectional top view, true shape of the section and development of the surface of the remaining part of the pyramid.
3. A cylinder 50mm diameter and 60mm long, is resting on its base on the ground. It is cut by a section plane perpendicular to the V.P.. the V.T. of which cuts the axis at a point 40mm from the base and makes an angle of 45° with the H.P. Draw its front view, sectional top view.
4. A triangular pyramid having base 40mm side and axis 50 mm long is lying on the H.P. on one its faces, with the axis parallel to the V.P. A section plane, parallel to the V.P. cuts the pyramid at the distance of 6mm from the axis. Draw its sectional front view and the top view.
5. A right circular cone diameter 54mm base and height 64mm, lies on one of its elements on H.P. with its axis parallel to V.P., A vertical section plane, parallel to the V.P. and 10mm away from the axis, cuts the cone. Draw the top view and sectional front view of the cut cone.
6. A cone base 75mm diameter and axis 80mm long is resting on its base on the H.P.
It is cut by a section plane perpendicular to the V.P., inclined at 45° to the H.P. and cutting the axis at a point 35mm from the apex. Draw its front view, sectional top view, sectional side view and true shape of the section.
7. A right regular pentagonal prism, base edge 30mm and height 70mm, is resting on H.P. with one of its base edges, such that its axis parallel to V.P. and inclined to H.P. at 30°. A horizontal section plane cuts the prism at a distance of 17mm from its top end. Draw its front view and sectional top view.
8. A right regular hexagonal prism, side of base 25mm and height 65mm, rest on an edge of its base on H.P., such that the rectangular face containing the base edge is inclined the H.P. at 30°. A section plane perpendicular to the H.P. and inclined to the V.P. at 45° cuts the prism such that the long edge farthest away from the V.P. is bisected. Draw the top view and sectional front view of the cut prism.
Development of surfaces
1. A cube of 30mm side is truncated as shown in fig. 1. Develop its lateral surface.
2. A cone is truncated as shown in fig.2. Develop its lateral surface.
3. The frustum of a square pyramid of side of base 36mm and height 40mm is resting on the H.P. with two parallel sides parallel to the V.P. Develop the slant surfaces of the solid fig. 3.
4. Develop the surface of the truncated cylinder as shown in fig. 4.
5. A triangular pyramid, side of base 30mm and height 60mm is truncated as shown in the fig 5. Develop the lateral surface of the solid if one of the sides of the base is perpendicular to the V.P.
6. A frustum of a cone of base diameter of 40mm, top diameter 20mm and height 40mm is cut off by a plane inclined at 30 degree with the horizontal and passes through the axis at 30mm above from the base. Develop the lateral surface of the remaining solid.
Intersection of surfaces
1. A vertical cylinder of 80mm diameter is completely penetrated by another cylinder of 60mm diameter, their axes bisecting each other at right angles. Draw their projections showing curve of penetration, assuming the axis of the penetrating cylinder to be parallel to the V.P.
2. A vertical cylinder of 75mm diameter is penetrated by another cylinder of 50mm diameter, the axis of which is parallel to both the HP. and V.P. The two axes are 9mm apart. Draw the projections showing curves of intersection.
3. A vertical cone, diameter of base 75mm and axis 100mm long, is completely penetrated by cylinder of 45mm diameter. The axis of the cylinder is parallel to the H.P. and the V.P. and intersects the axis of the cone at a point 20mm above the base.
Draw the projections of the solids showing curves of intersection.
4. A right cylinder of diameter 60mm and 90mm long, resting on its base on H.P., is penetrated by a square prism of 36mm base edge and 90mm long, such that the axes of the solids bisects each other at right angles. The faces of the prism are equally inclined to the V.P. Draw the projections of the solids showing curves of intersection.
5. A vertical cylinder of 60mm diameter has a square hole of 30mm sides cut through it. The axis of the hole is horizontal, parallel to the V.P. and 6mm away from the axis of the cylinder. The faces of the hole are equally inclined to the H.P. and V.P.
Draw the projections of the cylinder showing the hole in it.
Orthographic projection of solids
1. Draw the elevation, plan and end view of the given solids. As shown in figure enclosed.
2. Draw the sectional elevation, plan and end view of the given m/c parts.
UNIT-IV Pictorial Drawing
To make the drawing more understandable, several forms of 'one plane' conventional or projection drawings are used to supplement the orthographic drawings.
These one plane drawings, which can be easily understood by persons without any formal technical training are called pictorial drawings.
Methods of Projection: Pictorial drawings are projected on a single plane of projection by the methods of;
1. Axonometric Projection i. Isometric projection ii. Dimetric projection iii. Trimetric projection 2. Oblique Projection
3. Perspective Projection
Isometric Projection: The three axes form equal angles of 120° to the plane of projection, and only one scale is needed for measurements along each of the three axes.
Problems
1. A cube of 30mm edge is placed centrally on top of a cylindrical block of diameter 50mm and 60mm height. Draw the isometric projection of the solids.
2. A right regular hexagonal prism, edge of base 20mm and height 50mm, has a circular hole of diameter 20mm, drilled centrally through it, along its axis. Draw its isometric projection.
3. A cylindrical slab 70mm diameter and 40mm thick is surmounted by a cube of 35mm edge on the top of a cube, rest a square pyramid, altitude 35mm and side of base 20mm. The axes of the solid are in the same straight line. Draw the isometric projections of solids.
4. Orthographic views of different objects are given in fig. Draw the isometric projection for each of them.
UNIT-V
1. Draw the free hand sketch of the given m/c parts in fig.
2. Draw the single and double line plan of the given single storey building with electric wiring using symbols.
SINGLE STOREYED RESIDENTIAL BUILDINGS
Two Room Residential Buildings Problem
The line diagram of a Low Income Croup (LIG)house is shown in Some of the specifications are as follows:
Level of plinth above the ground = 500 mm Thickness of wall in plinth = 400mm Thickness of main walls = 300mm Thickness of partition wall between bath and W.C. = 100 mm Ceiling height for the rooms = 3000 mm Ceiling height for the verandah = 2700 mm
Doors : D = 1100 mm x 2100 mm; D1= 800 mm x 2000 mm Windows : W = 1500 mm x 1200 mm ; W1 = 1000 mm x 1200 mm Ventilators : V = 600 mm x 460 mm - Glazed
Roofing: Flat R.C.C. slab 125 mm thick
Parapet wall : 600 mm high and 100 mm thick above the rooms except the verandah.
Suitable foundation. D.P.C., flooring, lintels, sunshades, weathering course, steps, cooking platform in kitchen, shelves, shelves, etc. are to be provided.
Draw a suitable scale : a. The top view of the building at window still level.
b. The front view c. The section at IJ.
Problem
The line sketch of a Primary Health Centre with specifications is given below:
Specifications:
Level of the plinth above the ground : 600mm Height of the ceiling from floor : 3300 mm Thickness of the main walls : 300mm
Size of the cup-board : 1500 mm x 2100 mm
Doors : D = 1200 mm x 2100 mm ; D1= 1100 mm x 2100 mm.
Windows : W = 1650 mm x 1350 mm ; W1= 1100 mm x 1350 mm, Roofing : Reinforced cement concrete slab 12-5 mm thick.
Parapet wall : Height 600 mm and thickness 100 mm all-round.
Suitable foundation, steps, lintels, sunshades, flooring, D.P.C., weathering course, coping, etc are to be provided.
Draw the top view, front view and section at GH.
